用Python連接頂點以製作三角形

Question

我有一个双循环，看起来是这样的：

for i in xrange(len(myVVV)):
    for j in xrange(len(myVVV[i])):
        if myVVV[i][j]!= -1:
              s=i,j
              print(myVVV[i][j])
              print(s)

给出了：

^{pr2}$

我试着通过检查s代表网格上数字点位置的点来创建“三角形”，基本上我会连接点1，3，4，根据那里的位置形成一个三角形。有什么办法吗？在

我能读的东西？我正在创建一个刻面文件，这就是为什么我需要用顶点等三角形

Answer 1

这是我的代码，它很管用，对于以后的情况如果人们需要它请放心，希望它能有所帮助

MTri=0 #num of triangles 
numI=len(myVVV)#to check in range of i
numJ=len(myVVV[0])# to check in range of j
idsT=0 #for triangles
for i in xrange(len(myVVV)):
#print(myVVV[i])
for j in xrange(len(myVVV[i])):
    if myVVV[i][j] != -1:
        # line below check that if point has neighboring
        #points open to make a sqaure makeing it counter clockwise 
        #-> right, ^ up, <- left, v down (first case)
        if i-1 >= 0 and j+1 < numJ and myVVV[i][j+1] !=-1 and myVVV[i-1][j+1] !=-1 and myVVV[i-1][j] != -1:
            MTri= MTri +2#plus 2 since 2 traingles are made from a square
            A=myVVV[i][j]
            B=myVVV[i][j+1]
            C=myVVV[i-1][j]
            D=myVVV[i-1][j+1] 
            idsT=idsT+1
            #A,B,D create -> right ,^ up, back   _|
            #A,D,C create corner , <- left, back |/  
            print("create Triangle from points" + ' '+str(A)+' '+ str(B)+' '+ str(D))
            openwrite.write(str(idsT)+' '+str(A)+ ' '+str(B)+' '+str(D)+'\n')
            idsT=idsT+1
            openwrite.write(str(idsT)+' '+str(A)+ ' '+str(D)+' '+str(C)+'\n')
            print("create Triangle from points" + ' '+str(A)+' '+ str(D)+' '+ str(C))

        elif i-1 >= 0 and j+1 < numJ and myVVV[i][j+1] != -1 and myVVV[i-1][j+1] != -1:
            MTri= MTri+1#plus 1 
            A=myVVV[i][j]
            B=myVVV[i][j+1]# same index ,j shift -> one to the right
            C=myVVV[i-1][j+1] # index above, j shifted -> one to the right
            #A,B,C creates ->right,^ up, back  _|
            idsT=idsT+1
            openwrite.write(str(idsT)+' '+str(A)+ ' '+str(B)+' '+str(C)+'\n')
            print("create Triangle from points" + ' '+str(A)+' '+ str(B)+' '+ str(C))

        elif i-1 >= 0 and j+1 < numJ and myVVV[i][j+1] != -1 and myVVV[i-1][j] != -1:
            MTri= MTri+1#plus 1 
            A=myVVV[i][j]
            B=myVVV[i][j+1] #same index ,j shift -> one to the right
            C=myVVV[i-1][j]  #index above, same j position 
            #A,B,C creates ->right,corner, back  |_
            idsT=idsT+1
            openwrite.write(str(idsT)+' '+str(A)+ ' '+str(B)+' '+str(C)+'\n')
            print("create Triangle from points" + ' '+str(A)+' '+ str(B)+' '+ str(C))


        elif i-1 >= 0 and j+1 < numJ and myVVV[i-1][j+1] != -1 and myVVV[i-1][j] != -1:
            MTri= MTri+1#plus 1 
            A=myVVV[i][j]
            B=myVVV[i-1][j+1] # index above, j shifted -> one to the right
            C=myVVV[i-1][j]  #index above, same j position 
            #A,B,C creates corner right, left back  |/
            idsT=idsT+1
            openwrite.write(str(idsT)+' '+str(A)+ ' '+str(B)+' '+str(C)+'\n')
            print("create Triangle from points" + ' '+str(A)+' '+ str(B)+' '+ str(C))

Answer 2

你查过matplotlib了吗？这个答案here是用来画多边形的。

Answer 3

你可以试着反转坐标，从而证明两个点是对角的。你也可以用这个来检查对角线位置是否有一个点。然后，把它们组合起来，检查那个位置是否有一个点-如果有，你就有一个三角形。 例如

(0, 1) // One dot
Invert it to check for surrounding dots
(1, 0) // Dot found (inverted 0, 1)
(1, 1) // Combination - check if dot exists there. if true - triangle!

只是一个建议，可能会阻止一些想法到正确的道路：）我只是想到了这个，所以如果你找到一个洞，就放轻松点：p